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Math Help - Prove identity

  1. #1
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    Prove identity

    Prove that \arctan x+\arctan y\equiv \arctan\frac{x+y}{1-xy}.
    use this relationship to show that, if \arctan x+\arctan y+arctan z=\frac{\pi}{2} then xy+yz+xz=1
    I have done the first part:
    Let \arctan x=\theta and \arctan y=\phi
    then for LHS
    \equiv \theta +\phi
    \equiv \arctan(\tan(\theta+\phi))
    \equiv \arctan(\frac{tan \theta+\tan \phi}{1-\tan \theta \tan \phi})
    \equiv \arctan(\frac{x+y}{1-xy})\equiv RHS
    now my problem is the second part, after i subsitute \arctan(\frac{x+y}{1-xy}) into \arctan x+\arctan y+arctan z=\frac{\pi}{2} to get \arctan(\frac{x+y}{1-xy})+\arctan z=\frac{\pi}{2}, i don't know what to do.
    Thanks
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let \frac{x+y}{1-xy}=w

    Apply again the first identity for \arctan w+\arctan z and replace back w.
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  3. #3
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    thanks! i did that, and let arctan w=\omega and arctan z=\gamma
    so i had \omega+\gamma=\frac{\pi}{2}
    arctan(tan(\omega+\gamma)=\frac{\pi}{2}
    arctan(\frac{w+z}{1-wz})=\frac{\pi}{2}
    arctan(\frac{\frac{x+y}{1-xy}+z}{1-z(\frac{x+y}{1-xy})})=\frac{\pi}{2}
    arctan(\frac{x+y+z-xyz}{1-xy-xz-yz})=\frac{\pi}{2}
    now i'm stuck, tan\frac{\pi}{2} doesn't give a value
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  4. #4
    MHF Contributor red_dog's Avatar
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    If \arctan\frac{a}{b}=\frac{\pi}{2} then b must be 0.

    In this case 1-xy-xz-yz=0\Rightarrow xy+xz+yz=1

    Edited.
    Last edited by red_dog; July 30th 2009 at 12:45 AM.
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  5. #5
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    i thinking something along that line but didn't have any confirmation
    thanks!
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  6. #6
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    Quote Originally Posted by red_dog View Post
    If \arctan\frac{a}{b}=0 then b must be 0.

    In this case 1-xy-xz-yz=0\Rightarrow xy+xz+yz=1
    Whoops! I think you mean 'If \arctan\frac{a}{b}=\frac{\pi}{2} then b must be 0'.

    Grandad
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  7. #7
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by Grandad View Post
    Whoops! I think you mean 'If \arctan\frac{a}{b}=\frac{\pi}{2} then b must be 0'.

    Grandad
    Sorry! You're right! I've edited.
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